Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 3 - Differentiation - 3.2 Exercises - Page 174: 3

Answer

$f(x)$ is not differentiable on the interval [-1,1] since it is not differentiable at x = 0 due to cusp behavior.

Work Step by Step

For Rolle's Theorem to work, $f(x)$ must be differentiable over the closed interval. This is not true for $f(x) = \sqrt{(2-x^{2/3})^3}$ since it exhibits cusp-like behavior at x = 0. Because $f(x)$ is not differentiable at all points between the interval, Rolle's Theorem fails. The graph below depicts $f(x)$ and shows how $f(x)$ is not differentiable at x = 0.
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